Use the actual finite carrier Fin(5)={0,…,4} and its complete declared topology=[[], [0], [0, 2], [0, 2, 3], [0, 2, 3, 4], [0, 1, 2, 3, 4]], with subsets in the supplied order. Empty and whole sets, all finite intersections and all unions must belong to this finite topology. Connectedness means absence of a separation into two disjoint nonempty open sets covering this actual carrier, equivalently absence of a nontrivial clopen subset. It is not graph connectivity or a conclusion from selected tests. The empty-space convention here is explicitly the vacuous no-separation convention,not a nonempty ConnectedSpace type. Give (clopen_sets,nontrivial_clopen,connected?,exhaustive_complete_finite_topology), retaining the supplied topology order within both lists.
G001 — Use the actual finite carrier Fin(5)={0,…,4} and its complete declared topology=[[], [0], [0, 2], [0, 2, 3], [0, 2, 3, 4], [0, 1, 2, 3, 4]], with subsets in the supplied order. Empty and whole sets, all finite intersections and all unions must belong to this finite topology. Connectedness means absence of a separation into two disjoint nonempty open sets covering this actual carrier, equivalently absence of a nontrivial clopen subset. It is not graph connectivity or a conclusion from selected tests. The empty-space convention here is explicitly the vacuous no-separation convention,not a nonempty ConnectedSpace type. Give (clopen_sets,nontrivial_clopen,connected?,exhaustive_complete_finite_topology), retaining the supplied topology order within both lists.
Definition and scope
Use the original question domain, assumptions and cited MKO definitions. This item has second-context mathematical and bilingual review, with the model, answer and every arithmetic step independently recomputed from the original parameters; the source question remains a candidate.
unique empty/whole opens present;empty carrier allowed;first union failure=null;first intersection failure=null
Conditions: none
The actual n may be0,when empty and whole are the same open rather than two distinct sets.Check canonical subsets and all binary union/intersection obligations;the finite family then handles arbitrary unions.This data fixes the actual space,not a reference label.
The original member is open;it is clopen only if its complement is open.A separation obstruction also requires nonemptiness and properness.List all candidates in original topology order;one failed candidate cannot establish connectedness.
The original member is open;it is clopen only if its complement is open.A separation obstruction also requires nonemptiness and properness.List all candidates in original topology order;one failed candidate cannot establish connectedness.
The original member is open;it is clopen only if its complement is open.A separation obstruction also requires nonemptiness and properness.List all candidates in original topology order;one failed candidate cannot establish connectedness.
The original member is open;it is clopen only if its complement is open.A separation obstruction also requires nonemptiness and properness.List all candidates in original topology order;one failed candidate cannot establish connectedness.
The original member is open;it is clopen only if its complement is open.A separation obstruction also requires nonemptiness and properness.List all candidates in original topology order;one failed candidate cannot establish connectedness.
The original member is open;it is clopen only if its complement is open.A separation obstruction also requires nonemptiness and properness.List all candidates in original topology order;one failed candidate cannot establish connectedness.
ALL actual clopen sets=[[], [0, 1, 2, 3, 4]];ALL nontrivial clopen sets=[]
After
a nontrivial clopen S gives separation(S,X minus S);a separation by disjoint nonempty opens gives such a clopen S;actual connected iff no such set
Conditions: none
A proper nonempty clopen and its open complement separate the carrier.Conversely the two opens in a separation are complements,so either is nontrivial clopen.Exhausting the complete finite topology therefore excludes every separation.This is not connectivity of an unrelated graph.
The full substitutions and premise checks give the original answer; source identity,format and status are unchanged.
This question review and source binding
Review date: 2026-10-26
Original questions, answers, step expressions, numeric ASTs and historical records are retained. Source candidate labels remain historical and are not overwritten as formal proof.